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# The sum of all the digits of the integers from 18 to 21 incl

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Senior Manager
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The sum of all the digits of the integers from 18 to 21 incl  [#permalink]

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Updated on: 24 Jul 2014, 01:05
1
15
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Difficulty:

55% (hard)

Question Stats:

66% (02:26) correct 34% (02:26) wrong based on 364 sessions

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The sum of all the digits of the integers from 18 to 21 inclusive is 24 (1 + 8 + 1 + 9 + 2 + 0 + 2 + 1 = 24). What is the sum of all the digits of the integers from 0 to 99 inclusive?

A. 450
B. 810
C. 900
D. 1000
E. 1100

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Originally posted by shrive555 on 28 Oct 2010, 16:05.
Last edited by Bunuel on 24 Jul 2014, 01:05, edited 1 time in total.
Edited the question
Veritas Prep GMAT Instructor
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Re: The sum of all the digits of the integers from 18 to 21 incl  [#permalink]

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28 Oct 2010, 18:44
6
13
In the first 99 numbers, every digit appears 20 times. (Think 2 - It appears 10 times in units digit and 10 times in tens digit)

So all you need is (1 + 2 + 3 + ... + 9) x 20 = 45 x 20 = 900

Note: Remember 1 + 2 + 3 +...n = n(n+1)/2
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Re: The sum of all the digits of the integers from 18 to 21 incl  [#permalink]

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28 Oct 2010, 16:34
What we want is 0+(1+2+3+....+9)+(1+1+1+2+1+3...1+9)+.....(9+1+9+2..+9+9)

1+2+..+9=(9*10)/2=45
Similarly 11+12+...+19 will have 10 1s and again a sequence of (1+2+3+..+9). 20 series will be (2+0+2+1..2+9)=20+(1+2+3..+9)

Extrapolating this, we will get 45+(10+45)+(20+45)+....(90+45) = 45+(9*45)+(10+20+30+...+90)=450+10(1+2+..+9)=
450+10(45)=900.

Not sure if there is a better way to do this.
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28 Dec 2010, 11:07
1
Simple counting is what needed in this question
First of all we need to count how many 1 's are present from 0-99
i.e 1, 11-19, 21, 31, 41, 51, 61, 71, 81 ,91.... thus total of 20 1's are present
Thus total will be 1 * 20 = 20

Similarly there will be 20 2's and 20 3's ....... 20 9's

Therefore of each number will be 2*20 = 40, 3*20 = 60 ....... 9*20 = 180

Thus total of 20+40+60+80+100+120+140+160+180 = 900

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28 Dec 2010, 18:40
1
1
This is my method

Sum of the digits from 1 to 9 = Sum of consecutive numbers = first plus last divided by two X number of terms = 5 X 9 = 45

Number of times the digits from one to nine appear in the sequence from 1 to 99 = 10

First part of the answer = 45 X 10 = 450

Number of times the number one (two..nine) appears in the sequence from 1 to 99 in the units digit = 10
1 X 10 = 10
2 X 10 = 20
.....
Sum of consecutive numbers = First Plus Last / 2 X Number of terms

90 + 10 = 100 / 2 = 50 X 9 = 450 Second part of the answer

Final answer 450 + 450 = 900

It worked for me...

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28 Dec 2010, 19:52
1
1+2+3+4+5+6+7+8+9
(1+0)+(1+1)+(1+2)+(1+3)+(1+4)+(1+5)+(1+6)+(1+7)+(1+8)+(1+9)
(2+0)+(2+1)+(2+2)+(2+3)+(2+4)+(2+5)+(2+6)+(2+7)+(2+8)+(2+9)
and so on..

You don't have to write the above in the test... just quickly visualize it. What do we see here?

1+2+... +9, which adds to 45 appears 10 times
=> 45*10 = 450 -------------- (1)

also, we have 10 1's, 10 2's and so on
=> 10 (1 + 2 +...+ 9)
=> 10 * 45 = 450 ------------- (2)

(1) + (2) = 900

Hope this helps!
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24 Jan 2011, 11:50
1

think of all numbers between 0-99 as double-digit numbers(1=01, 2=02 and so on)

sum of unit digits
digits 1-9 repeat 10 times (01 to 09, 11 to 19 .... 91 to 99)

sum of unit digits $$= 10*[\frac{9}{2}*(1+9)] = 450$$

sum of tens digits
digits 1-9 repeat 10 times (10 to 19, 20 to 29 .... 90 to 99)

sum of unit digits $$= 10*[\frac{9}{2}*(1+9)] = 450$$

total = 450 + 450 = 900

Ans: C
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24 Jan 2011, 20:06
6
shrive555 wrote:
The sum of all the digits of the integers from 18 to 21 inclusive is 24 (1+8 + 1+9 + 2+0 + 2+1 = 24). What is the sum of all the digits of the integers from 0 to 99 inclusive?

450
810
900
1000
1100

0 to 99 could be considered 100 two digit numbers(00, 01, 02....99) since when 0 is added, it doesn't change the sum. To write these 100 numbers, 200 digits will be used. Since there are 10 digits (0 to 9), each digit will appear 20 times.

So total sum = 20*0 + 20*1 + 20*2 + 20*3 +.....20*9 = 20*9*10/2 = 900
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03 May 2011, 19:45
0+1+2+...+9 = x = 45

now we have,
unit's digit sum from 0-99 = 10 * x
ten's digit sum from 0-99 = 10* x

hence total is 20*x = 20 * (45) = 900. Hence C.
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07 May 2011, 08:44
we can see the pattern here

units digit : 1,2,3,4,5,6,7,8,9 repeated 10 times

=> unit digit sum = 10*(1+2+3+4+5+6+7+8+9)

tens digit : 1 repeated 10 times, 2 repeated 10 times...9 repeated 10 times

=> tens digit sum = 10*(1+2+3+4+5+6+7+8+9)

so their sum = 20*(1+2+3+4+5+6+7+8+9) = 900

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Re: The sum of all the digits of the integers from 18 to 21 incl  [#permalink]

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03 Jan 2014, 04:57
3
1
shrive555 wrote:
The sum of all the digits of the integers from 18 to 21 inclusive is 24 (1+8+1+9+2+0+ 2+1=24). What is the sum of all the digits of the integers from 0 to 99 inclusive?

A. 450
B. 810
C. 900
D. 1000
E. 1100

1+2+3+4...+9 sum of all digits is 45

Each digit will appear ten times in the units digit = 10*45= 450
Each digit will appear once in the tens digit = 1*45*10 = 450 (The ten is cause its in the tens digit)
Total= 900

Hope it helps

Gimme Kuddos

Cheers
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Re: The sum of all the digits of the integers from 18 to 21 incl  [#permalink]

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03 Jan 2014, 05:12
1
Sum from 0 - 9 = 45
Sum from 10 - 19 = (1x10 + 45) = 55
Sum from 20 - 29 = (2x10 + 45) = 65
.
.
.
Sum from 90 - 99 = (9x10 + 45) = 135

Total sum = 55 + 65 ..... 135 = 10(45+135)/2 = 900
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Re: The sum of all the digits of the integers from 18 to 21 incl  [#permalink]

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24 Feb 2014, 21:24
1
0 to 9 in units place appear 10 times, so 10x0 + 10x1 + ............ + 10x9 = 450
0 to 9 in tens place appear 10 times, so 10x0 + 10x1 + ............ + 10x9 = 450

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The sum of all the digits of the integers from 18 to 21 incl  [#permalink]

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17 Jun 2016, 08:08
shrive555 wrote:
The sum of all the digits of the integers from 18 to 21 inclusive is 24 (1 + 8 + 1 + 9 + 2 + 0 + 2 + 1 = 24). What is the sum of all the digits of the integers from 0 to 99 inclusive?

A. 450
B. 810
C. 900
D. 1000
E. 1100

Many of the over-mentioned require a good number sense, which I do not posses, so I'll show you my "layman" approach

We want the sum of the digits from 0 to 99, so I approximated:
0-9 -> 45 -> (9+0)*10/2
40-49 -> 85 (13+4)*10/2
90-99 -> 135 (18+9)*10/2

We can see at a glance that the "weight" goes up as the numbers go up (meaning the difference between 85 and 45 is 40, while 135-85 is 50, this means that the second part of this sequence carries more weight for our result), so we know that the final answer has to be more than 850 (85*10) but close to it, and that's just 900: the answer is C.

Kudos if you like.

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Re: The sum of all the digits of the integers from 18 to 21 incl  [#permalink]

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25 Aug 2018, 03:21
In the first 99 numbers, every digit appears 20 times. (Think 2 - It appears 10 times in units digit and 10 times in tens digit)

So all you need is (1 + 2 + 3 + ... + 9) x 20 = 45 x 20 = 900

Note: Remember 1 + 2 + 3 +...n = n(n+1)/2

VeritasKarishma, your solution to this is great, but do you not think that there must be reason to give 18 to 21 = 24. Is there a possibility to use this and reach to the answer earlier?
Usually Questions mentions "such as" or "for example" if they add info just to make us understand the question better!!!

Since this question is from Magoosh, mikemcgarry, can you provide the official solution?
Re: The sum of all the digits of the integers from 18 to 21 incl &nbs [#permalink] 25 Aug 2018, 03:21
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